Argument of complex number - Practice Questions & MCQ

If a complex number z = x + iy is represented by a point P in the Argand plane and OP forms some angle with a positive x-axis, let's denote it with ?, then ? is called the argument of z.

$\begin{aligned} & \tan \theta=\frac{\mathrm{PM}}{\mathrm{OM}} \\ & \tan \theta=\frac{\mathrm{y}}{\mathrm{x}}=\frac{\operatorname{Im}(\mathrm{z})}{\operatorname{Re}(\mathrm{z})} \Rightarrow \theta=\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}} \\ & \arg (\mathrm{z})=\theta=\tan ^{-1} \frac{\mathrm{y}}{\mathrm{x}}\end{aligned}$

If ? lies between -? < ? ≤ ?, then ? is called the principal argument. The value of the argument differs depending on which quadrant point (x,y) lies.

If it lies in 1^st quadrant then it is ? (acute angle)

If the point lies in 2^nd quadrant, then  $\arg (z)=\theta=\pi-\tan ^{-1} \frac{y}{|x|}$ 

So it will be an obtuse +ve angle

If the point lies in lies in 3^rd quadrant then  $\arg (z)=\theta=-\pi+\tan ^{-1} \frac{y}{x}$

It will be an obtuse -ve angle

If the point lies in 4^th quadrant then  $\arg (z)=\theta=-\tan ^{-1} \frac{|y|}{x}$

It will be a -ve acute angle

Note:

If $\arg (\mathrm{z})=\frac{\pi}{2}$ or $-\frac{\pi}{2}, \mathrm{z}$ is purely imaginary.
If $\arg (\mathrm{z})=0$ or $\pi, \mathrm{z}$ is purely real.